\(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

\(x^3-5x^2+x-5=0\)

\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

a) x² - 3x - x(x + 2) = 2

=> x² - 3x - x² - 2x = 2

=> -5x = 2

=> x = -2/5

b) 5x³ - 3x² + 10x - 6 = 0

=>x²(5x - 3) + 2(5x - 3) = 0

=> (x² + 2)(5x - 3) = 0

=> \(\orbr{\begin{cases}x^2+2=0\\5x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x^2=-2\left(ktm\right)\\5x=3\end{cases}}\)

=> x = 3/5

\(a,x^2-3x-x\cdot\left(x+2\right)=2\)

\(x^2-3x-x^2-2x=2\)

\(-5x=2\)

\(x=-\frac{2}{5}\)

\(b,5x^3-3x^2+10x-6=0\)

\(5x\cdot\left(x^2+2\right)-3\cdot\left(x^2+2\right)=0\)

\(\left(x^2+2\right)\cdot\left(5x-3\right)=0\)

\(\hept{\begin{cases}x^2+2=0\\5x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x\notin\varnothing\\x=\frac{3}{5}\end{cases}}}\)

Vậy......

x=0 nha

x²-10x+16=0

x²-2.5x+25-9=0

x²-2.5x+25  =9

(x-5)²          =3²

x-5             =3

x                =8

Mấy bài này dễ lắm,bn làm tương tự nha(câu nào không làm theo hằng đẳng thức được thì tách

a,

Ta có

x²+5x+6=x²+6x-x+6=0

=>x(x-1)+6(x-1)=0

=>(x+6)(x-1)=0

=> x=-6 hoặc x=1

KL

b,x²-10x+16=0

=>x²-8x-2x+16=0

=>x²-2x+16-8x=0

=>x(x-2)-8(x-2)=0

=>(x-8)(x-2)=0

=>x=8 hoặc x=2

KL

c,2x²+7x+3=0

=>2x²+6x+x+3=0

=>2x(x+3)+(x+3)=0

=>(x+3)(2x+1)=0

=>x=-1/2 hoặc x=-3

KL

d,

x²-10x+21=0

=>x²-3x-7x+21=0

=>x(x-3)-7(x-3)=0

=>(x-7)(x-3)=0

=>x=7 hoặc x=3

KL

e,x²-2x-3=0

=>(x²-2x+1)-4=0

=>(x-1)²-2²=0

=>(x-1+2)(x-1-2)=0

=>(x+1)(x-3)=0

=> x=-1 hoặc x=3

KL

f,

x³-7x-6=0

=>x³+x²-x²-x-6x-6=0

=>x²(x+1)-x(x+1)-6(x+1)=0

=>(x²-x-6)(x+1)=0

=>(x²-3x+2x-6)(x+1)=0

=>[x(x+2)-3(x+2)](x+1)=0

=>(x+1)(x+2)(x-3)=0

=>x=-1 hoặc x=-2 hoặc x=3

KL

Bài 3:

1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)

\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy.......................

2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)

Vậy........................

3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy............................

4, 5 tương tự nhé bn!

bài 3

1 (x-1)(x+2)+5x-5=0

=>(x-1)(x+2)+(5x-5)=o

=>(x-1)(x+2)+5(x-1)=0

=>(x-1)(x+2+5)=0

=>(x-1)(x+7)=0

=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

vậy x=1 hoặc x=-7

2. (3x+5)(x-3)-6x-10=0

=>(3x+5)(x-3)-(6x+10)=0

=>(3x+5)(x-3)-2(3x+5)=0

=>(3x+5)(x-3-2)=0

=>(3x+5)(x-5)=0

=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)

Bài 1:

a) \(x^2-10x=-25\)

\(\Rightarrow x^2-10+25=0\)

\(\Rightarrow x^2-2.x.5+5^2=0\)

\(\Rightarrow\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=0+5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

b) \(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow x.\left(x-2\right)-3.\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right).\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+2\\x=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{2;3\right\}.\)

Chúc bạn học tốt!

a) $x^2-10x=-25$

$\mathrm{<=>\; \backslash (x^2-10x+25=0\backslash )}$

$\mathrm{<=>\; \backslash (\backslash left(x-5\backslash right)^2\backslash )}$=0

=>\(x-5=0\)

=>\(x=5\)

b) $x^2-5x+6=0$

<=> (x-2)(x-3)=0

=>x-2=0 hoặcx-3=0

=>x=2 hoặc x=3

c) rút gọn rồi phân tích đa thức thành nhân tử

a.

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2\ge0\\3x^2-17x+4=\left(3x-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\3x^2-17x+4=9x^2-12x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\6x^2+5x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\left[{}\begin{matrix}x=0< \dfrac{2}{3}\left(loại\right)\\x=-\dfrac{5}{6}< \dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

b.

ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\)

Đặt \(\sqrt{x^2-5x+4}=t\ge0\Leftrightarrow x^2-5x=t^2-4\)

\(\Rightarrow2x^2-10x=2t^2-8\)

Phương trình trở thành:

\(2t^2-8-3t+6=0\)

\(\Leftrightarrow2t^2-3t-2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{1}{2}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-5x+4}=2\)

\(\Leftrightarrow x^2-5x=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

3x+18y=3(x+6y)

b, \(15\left(x+3\right)+20x\left(x+8\right)=15x+45+20x^2+160x\)

\(=20x^2+175x+45=...\)

c, \(6\left(x-9\right)-3x\left(y-x\right)=6x-54-3xy+3x^2\)

d, \(2xy+10x^2-x\) không phân tích được nữa nhé

e, \(4ab^2-28a+16b\)không phân tích được nữa nhé

g, \(a\left(a+b\right)=ab\left(a+b\right)< =>\left(a+b\right)\left(a-ab\right)=0< =>\left(a+b\right)a\left(1-b\right)=0\)

h, \(30a^2+6a-6=\left(\sqrt{30}a\right)^2+2.\sqrt{30}.\frac{3}{\sqrt{30}}+\frac{3}{10}-\frac{63}{10}\)

\(=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}\right)^2-\sqrt{\frac{63}{10}}^2=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}-\sqrt{\frac{63}{10}}\right)\left(\sqrt{30}a+\frac{3}{\sqrt{30}}+\sqrt{\frac{63}{10}}\right)\)

a) Ta có: 5x(12x-7)-6(10x²+3) = 0

\(\Leftrightarrow\) 60x²-35x-60x²-18 = 0

\(\Leftrightarrow\) -35x = 18

\(\Leftrightarrow\) x = \(-\dfrac{18}{35}\)